poj 3660 Cow Contest Flyod

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5989		Accepted: 3234

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

题意：是给你n个人，m组关系
每次关系输入A,B。表示A比B屌
思路：Flyod跑一发，然后判断是否这个人和其他人的关系都已经确定，如果都已经确定，那么就直接ans++就好了！

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 105
#define eps 1e-9
const int inf=0x7fffffff;   //无限大
int g[maxn][maxn];
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        int a,b;
        for(int i=0;i<m;i++)
        {
            cin>>a>>b;
            g[a][b]=1;
        }
        for(int k=1;k<=n;k++)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(g[i][k]&&g[k][j])
                        g[i][j]=1;
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            int flag=1;
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                    continue;
                if(g[i][j]==0&&g[j][i]==0)
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
                ans++;
        }
        cout<<ans<<endl;
    }
}